Only A "Genius" Can Solve - Each Letter Is A Number

What value is each letter? This is an incredible math problem with just a single solution! Can you figure it out?
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[@sajadmahmood9509]

Usually I don’t solve the puzzles but rather just watch the video and learn to solve them. But once I saw the thumbnail of this video I grabbed a pen and paper and started going at it after like 5 minutes I got lost in the carry overs and decided to give up and watch the video instead. When u said “I strongly suggest that you try solving this one” I was determined to solve it and started fresh and got it in like 10 minutes ,I was so proud of myself! Thank you mr.talwalker for all the videos and especially for this amazing puzzle.

[@Miaumiau3333]

good job :)

[@sajadmahmood9509]

MiauMiau :3 Thx

[@coproski6276]

Same exactly thing for me, after 10 minutes I though, meh I will just watch the video like I always do, but the "I strongly suggest" part encouraged me enough to solve the puzzle. After 15 more minutes I could get the result. Congratz Sajad to not give up. :D

[@uncann130]

Sajad Mahmood This was what I did too!

[@cocoguerra2897]

Just did the same

[@euanocurry2144]

This feels more like an unsubtle hint about your financial situation more than a video about maths

[@jbtechcon7434]

Next week's puzzle: PATREON + SPONSORS = GROCERIES

[@thewatcherinthecloud]

Solution is erroneous because S=0 (N+S=S on the first digits, so S has to be 0), and you can't carry make 0+1=GR, unless you set different parameters for the problem. Also, you have 11 different letters, meaning you would have more than 11 digits, meaning we can't solve this problem on base 10.

[@jbtechcon7434]

thewatcherinthecloud N+S=S would imply that N=0 not S=0. But anyway, I didn't say we were using base 10, and I didn't say the letters represent different numbers.

[@shadann3650]

Can someone answer this question🙏 Find the pattern to this sequence: -1,-1,-1,+1,+1,+1,-1,-1,-1,+1,+1,+1...

[@davidjames1684]

I took it a step farther and said there are 13 letters so we need at least 13 distinct digits.

[@josephmercurio4759]

This was a bonus problem at the math summer camp I worked at this summer! I am proud to say that most of the kids, who were 13 or 14 years old, figured it out!

[@andywong3095]

Glad to hear it !!!! , It, REALLY, make me feel young. Ha, I am 14 years old now. Thanks for that Daydreaming.

[@XxStar1ightXx]

I had these type of problems in my math class when I was 11. Basically everyone did it, but it took a long time 😅

[@florianzeau4527]

"Send money" is there a hidden message? 😂

[@maxboo7995]

Florian Zeau only a genius can solve this, hmmmmmm...........

[@PeroG]

@@maxboo7995 its send more money XD

[@recarras]

I dont think it is hidden at all.

[@Reg_al19]

@@recarras it is, it's known as cryptarithmetic problems

[@1o1ho1up]

Hey this is pressure walker... I love the subtitles

[@OneWeirdDude]

That's one good reason why I try to fix them. But yeah, that and "pressure locker" are funny. In all fairness, I've never heard of "Presh" or "Talwalkar" as names before. Can you blame KZbin's artificial intelligence?

[@sairamniranjan3908]

@@OneWeirdDude Presh Talwalkar is probably an Indian name.

[@atharvas4399]

talkwalkar is. Presh... i dont think so.

[@simplebutpowerful]

Is Presh short for something?

[@sairamniranjan3908]

@@simplebutpowerful yeah, Presh is a nickname for Prashant.

[@JekyllsOtherHalf]

I noticed that you can swap letters in a single column and the sum will still work. If you swap the N and R around, you get SERD + MONE = MONEY, which suggests that SERD = MONE × 9 + Y. Solved for Y = 0, then increased Y til I had unique values.

[@arthshah3378]

appreciate

[@parthpandya6412]

Wow

[@ghimbos]

how did you solve SERD = MONE x 9 ? Is this is any simpler that solving the original one?

[@StarTheTripleDevil]

@@ghimbos SERD + MONE = MONEY SERD = MONEY - MONE SERD = 10 × MONE + Y - MONE SERD = 9 × MONEY + Y

[@fresh5899]

what the f-

[@fadetounforgiven]

Man, I was told about this puzzle by my mum many, many years ago, maybe thirty or so, and she had heard about it many, many years before. The "story" was that a son went and study English in England and when he run out of money, he sent a letter/telegram to his parents with only that text "send more money". So his parents wrote him back with that puzzle, and the son would have to solve it if he wanted to get that money. And this is Northwest Spain... so that thing is old but brings back some memories.

[@gonshi9]

Hola xD, veo que no soy el único español

[@fadetounforgiven]

Como dice al final del vídeo, así "a lo humilde", estos problemas inspiran a millones alrededor del mundo... No sé si tanto pero era de esperar que no fueses el único.

[@andywong3095]

Fade or Fake, depend on Knowledge. Knowledge Not Equal to IQ. (if YOU knew, to consider A twin as A single object, it will be easy). I may add, A twin, is a pair of OBJECT.

[@atriyakoller136]

I like the Russian version of a student who was in need of money. He wrote a telegram "student", which was supposed to mean (I'll give the transliteration and translation) "srochno trebuetsya uima deneg est' nechego tochka" which means "I urgently need a huge amount of money, there's nothing to eat, period". It's not mathematical but it's funny

[@eliasfer4009]

@@gonshi9 Yo también lo soy Viva España!

[@sairamniranjan3908]

Kinda tricky, but really satisfying problem. Thank you, Presh, for making this video.

[@iamrepairmanman]

Starting from M=1 (which is the max carryover) that forces S to be 9 or 8, and O has to be 0. Seeing as O is 0, that means there is no carry over, so S is 9. This means there is a carry from the tens place, and E=N-1. In order to get one less(and a carry) from addition, you need to add 9, but 9 is already used, making R=8, and D+E needs a carry. We only have 7, 6, 5, 4, 3, and 2 left to use. Setting D to 7, and E to 5 makes Y=2, and N=6, this fulfills all our requirements. Our equation is therefore: 9567+1085=10652

[@shivaargula4735]

This was my logic process too, although it took me 10+ min to get to that point. A good 5 min was figuring out where to start, though.

[@nettekal8190]

Exactly the same step as me.. took me 5 minutes, then i check the video, then read your comment.. have a nice day bro

[@nanashi_74_]

wowowowow awesome

[@PratapSingh-is4eb]

I did the exact same thing bro.....cheers 🍻🍻🍻🍻🍺

[@ralphy1054]

Nice explanation!

[@SreenikethanI]

Although I didn't do it on paper, I created a program to iterate from 0 to 9 for all the 8 letters, and got 25 solutions. But then I realized that M ≠ 0, so there was only one left. Also tried creating a small Excel sheet to verify any result. Yeah, I didn't do the math when I'm supposed to :D

[@ykkhatri]

Sreenikethan sir can you please guide on how to make such a program, by replying to this comment or by perhaps making a video about it.

[@SreenikethanI]

Okay, I'll try making a video! :) By the way, please don't call me sir and all, I'm surely younger than you!

[@SreenikethanI]

It's basically nested For loops iterating through 0-9 for each of the letters (S, E, N, D, M, O, R, Y) and verifying whether the equation is satisfied, as well as whether M ≠ 0. As there are 10 values (0-9) for each letter, there will be a total of 100,000,000 possible combinations for values of letters possible.

[@ykkhatri]

Sreenikethan I I am getting getting confused in how to keep the letters distinct.

[@ykkhatri]

Sreenikethan I a video will be great....Just Pressed the bell on your channel

[@Fanaprimo]

I solved this one 55 years ago. Still hice memories about it.

[@apatriot6421]

You dead?

[@brightwriter]

Similar to the puzzle, identify the digits represented by the letters, in ONE + TWO + FOUR = SEVEN, which does not give a unique solution, but add the constraint that if you multiply your WOES by 5 you get SORER, it does. I've seen the solution somewhere, but I can't remember it.

[@jonnysanimecorner196]

"Did you figure it out?" took me over 90 min, but yes I did

[@teambellavsteamalice]

Nice problem! And I got it ☺️. I almost gave up with just finding M, but after your encouragement I tried and saw that S is either 8 or 9 and O = 0. I worked with vectors of remaining possibilities (sudoku style), like E ={2,3,4,5,6}, N={3-7} and found R was also {9,8}. I actually found D=7 before I saw S = 9 and R = 8, but then it was just one more step to figure out which E={5,6} had to be right.

[@shivaargula4735]

I gave up and played the video, and your additional message appeared encouraging me to try to solve it. I paused the video and tried again and came to the solution! I'm glad I did, that was a fun puzzle. The trick was you said they are all "unique" numbers, which I didn't catch at first! Took me about 15 min. I solved most of it logically, but towards the end I tried out the possibilities for the remaining numbers (6 possibilities) and found the unique solution! Thanks for the encouragement!

[@indigoziona]

I really enjoyed this one: working it out, working out the reasoning to show I was correct, and watching your explanation also!

[@philipcollier4883]

Reminds me of that book "Sidways arithmatic from wayside school" almost the entire book had word equasions in it. Hurt my head in the elementry school. Great vid!

[@roquedefrutos8667]

1:56 This was quite funny to me: "Since 999+999

[@markorezic3131]

When your patreon supporters join the math meanies

[@djgulston]

I was able to solve M, O, S and R, but I was stuck on N, E, D and Y. I managed to obtain the equations D + E = MY and E + 1 = N beforehand and didn't know how to move forward from there, so I watched up until you mentioned D + E >= 12 and I managed to solve the remaining 4 letters. This was a fun exercise, I hope to see a similar one in the future.

[@rinkeborg]

yes that was the trickyest part. I figured that E could only be 5 or 6, since all other possibilities would either create values already occupied by other letters, or wouldn't create a carry over of 1. And when you have only two possibilities left for E, its basically just trial and error.

[@user-ny7di4sx6k]

One of the few exercises in this channel I solved by myself.

[@odinfeidje-baug7467]

It's so fitting that the roundest letter gets the roundest number.

[@Mikeharuki]

It took me about 40mins to solve and it was such a blissful feeling when I finally found the solution. It was much less logically structured compared to your explanation but still ;) Thank you so much for sharing this

[@AREmrys]

I used a slightly simpler thinking to work out the value of R: As it is clear that E+1 = N, we can only get back from N to E by subtracting 1, which means adding 9 in this modulo 10 system. As 9 is already the definite value of S, R can only be 8 with a carryover from the far right column.

[@Refreshhhh18]

Yes used same😄

[@blackphnix264]

So much satisfaction when you figure it out alone 😁

[@Chloe-sb9tt]

can't relate ✨

[@Shubham07M]

yeah

[@EnerJetix]

I thought I was good in Math until I watched this channel

[@damogetscarried766]

This question was more about logical reasoning rather than mathematical ability.

[@carolyn8740]

There is a treat on a table on the other side of a room, and with every step you take you half the distance separating you and the treat (because you're awesome), will you ever get to eat the treat?

[@borizkyjevi304]

@@damogetscarried766 bro it's a joke

[@marcusbressan28]

@@carolyn8740 no

[@SomeGuy712x]

I figured it out, using pretty much the same logic you did. Took me maybe 5-10 minutes and I typed the logic notes I was taking into Notepad while working this out.

[@thewatcherinthecloud]

Knowing that E+1=N was the greatest triumph of my day.

[@j-k-l4756]

Oh the feels, did so too

[@kunalbarapatre7197]

Not to brag, but this question was asked in an exam conducted in my state, back in early 2000s, to 12-13 year olds (8th grade students). And the students are expected to solve this question in 5-10 minutes. It brings me real happiness that something from my childhood finally came up on your channel! 😃

[@yolostratz]

What happens if you reverse the puzzle (like the letters are represented by numbers, instead numbers are represented by letters), will it be harder to solve, 'cuz there are 26 letters in the alphabet, so that means more possibilities right?

[@stereotypicalblobby]

Before watching solution: when adding, the most overflow you can get by adding two single-digit numbers is 1 (as in a 1 gets sent to the next row over). Because of this, m has to equal 1 since two single-digit numbers are being added together to form a two-digit number. With this, we can also determine that s can either equal 8 or 9, depending on whether or not there is overflow from e+o. Looking at the possibilities, if s is equal to 8, then we would have 8 + 1 plus 1 from overflow from e+o to give us the two-digit number 10, making o equal to 0. However, if this were true, that would make e+o some single-digit number plus 0 to make a two-digit number, which isn't possible (even with possible overflow from n+r, that would mean both n and o are 0, which isn't allowed). Thus, s must equal 9. Since s is 9 and m is 1, adding those together would give us either 10 or 11 with overflow from e+o, meaning o is either 0 or 1. However, since m is already 1, o must be 0. (To solve e I go off on a few different paths) because each letter represents a different digit, there must be overflow from n+r into e+o to keep e and n from being the same. Thus, n is 1 greater than e. Because we see r being added to some number to get 10 plus that number - 1, r would have to be 9 (8+9 = 17, 5+9 = 14, and so on). However, since s is already 9, r must be 8 and there must be overflow from d+e. Because r is 8, e cannot be 7 because n would then also be 8, which isn't allowed. Also, since there has to be overflow from d+e and o is already 0, e must be greater than 3 because d cannot be 8 or 9 and d being 7 with e being 3 would mean y is 0 (which it cannot). Therefore, e must be between 4 and 6, inclusive. Now I have to go through the three possibilities for e: 4, 5, or 6. If e is 4, then d must be at least 7 to make y greater than 0. However, since d cannot be 8 or 9, d would have to be 7, which would make y 1, which would conflict with m being 1, so e cannot be 4. If e is 5, then d would still have to be 7 because otherwise it would cause a repeat of the previous scenario. If e is 5, d is 7, y is 2, and n is 6, which would give us 9567+1085=10652, which checks out mathematically. However, I should still check to make sure that e cannot be 6. If e is 6, d would still have to be 7, because d being 6 would lead to both e and d being 6, which isn't allowed. 7+6 gives us 13, making y 3, which is allowed. However, if e is 6, then n would have to be 7, but d is already 7, so e cannot be 6. Thus e is 5, d is 7, n is 6, and y is 2. Final solution: s=9, e=5, n=6, d=7, m=1, o=0, r=8, and y=2.

[@keziahatten8495]

Me the entire video: "Child, what!!!" Goes back aqain replay again.

[@davidb5205]

Fun puzzle! Took me half an hour to complete. I got tripped up at a few spots; glad I worked it out. *My solution:* 'M' is obviously 1 because 1 is the max overflow and it's not 0. 'S' has to be either 8 (with a 1 overflow from the previous column) or 9, either way 'O'=0. Assume 'S'=9. Looking at the middle column, since 'E'+0='N' leads to 'E' and 'N' having the same value, there MUST be an overflow 1 from the previous column. Then you get *'E'+1='N' and 'N'+'R'='E'+10* from the middle two columns. I got stuck because the PROBLEM is that all solutions to these pair of equations give you 'R'=9 which is already taken by 'S'. This is fixed by assuming overflow in the 'NRE' column as well. So the pair of equations become *'E'+1='N' and 'N'+'R'+1='E'+10* which makes 'R'=8. That works! Moving on, since 'E'+1='N', 'E' and 'N' must be a subsequent pair (2,3)(3,4)(4,5)(5,6) or (6,7). Only one of these pairs has an 'E' big enough to create overflow when added to 'D' and also big enough such that 'Y' is a unique number. That unique pair is (5,6). *Answer:* 'M'=1; 'O'=0; 'S'=9; 'R'=8; 'E'=5; 'N'=6; 'D'=7; 'Y'=2 *9567 + 1085 = 10652*

[@darpanshrestha2190]

Engineering problem 7th sem artificial intelligence subject 😂😂😂😂

[@Adarsh-mn7pl]

LoL, kal exam hai😂😂😂

[@Reg_al19]

Mera class 8th is hai yeh question

[@mortiphasm5597]

This was a fun problem. It was fairly easy to work out that M was 1, O was 0 and S was 9, as well as that N=E+1, but the D and the Y made things tricky.

[@DonMaguz]

E, N and D are the keys to the uniqueness of this puzzle. A very quick one, but a nice one nonetheless.

[@DonMaguz]

Took you 30 minutes? Ok... maybe if you tried to prove its unique solution. Will proceed watching now...

[@DonMaguz]

Ohh I see, the way I solved it faster was by having the digit list on clear sight and crossing down the used ones. That way you have nowhere to move once you cross down the 8. Took me around 7 minutes by the way, not a record by any means but that's why it surprised me that it took Presh half an hour.

[@nicholasleclerc1583]

David Martínez “ *A* very unique...” ; P

[@sairamniranjan3908]

@@DonMaguz it probably took Presh about ten minutes to solve this problem, but another twenty minutes to devise a foolproof method to present the solution. You know, the intricate animation of numbers to make it easier for us to understand and enjoy the problem.

[@paulpeeters]

I started in the same way (numbering columns from right to left as 1 to 5) but continued with a different logical reasoning : M can only be equal to 1 and we must have carry over from column 4 (==> M=1, co4=1). As we require carry over from column 4, we can conclude that S can only be 8, with carry over from column 3 required, or 9, with no carry over from column 3 required and because O cannot be 1 as 1 is already assigned to M, O can only be 0 (==> M=1, co4=1, S=8/9, O=0). As we replace the O in column 3 with 0 as well, we can conclude that there must be a carry over from column 2 to have E N and we can temporarily set E = x, N = x+1 (==> M=1, co4=1, S=8/9, O=0, co2=1). We substitute E and N also with x and x+1 in columns 2 and 1. In column 2, because the last digit of x+1+R is x, we can decide that R can only be 8 if we have carry over from column 1, or 9 if we don’t have carry over from column 1 (==> M=1, cy4=1, S=8/9, O=0, R=8/9). A carry over in column 3 would only be possible if E were equal to 9 which is impossible because 9 is already assigned to S or R. As S can only be equal to 8 when we have a carry over from column 3, S must be equal to 9 and as a result R should be equal to 8 (==> M=1, cy4=1, S=9, O=0, R=8, co3=0). Because R is 8, we know that we need carry over from column 1 (==> M=1, cy4=1, S=9, O=0, R=8, co3=0, co1=1). To have carry over in column 1 while not using 8 or 9 for D nor for E and the last digit of their sum not resulting in 0 or 1 for Y, we need D and E to be between 5 and 7. Because E = x and N = x+1, we can conclude that E must be 5, that N must be 6 and that D must be 7 (E could not be 6 because that would mean that N would be 7 and D would be 5 resulting in the last digit of sum D+E equal to 1). With E = 5 and D = 7, we also solved the last letter Y with will be 2 (==> M=1, cy4=1, S=9, O=0, R=8, co3=0, co1=1, E=5, N=6, D=7, Y=2). This resolves the puzzle.

[@BaoNguyen-ek4lu]

The first time i can say yes when he askes "Can you figure it out? "

[@zunkman1]

I solved it successfully and then heard your explanation, which got me so lost that I was doubting my own answer.

[@crappyj7603]

It took me 6,227,020,800 tries to find this one

[@machimuse7526]

I’m starting to see a general pattern in solving these types of puzzles but I can’t quite articulate it. Something like (in no particular order): 1) Work backwards 2) Start plugging in the higher digits 3) Look for logical constraints and stack them onto each other to further constraint the problem 4) Anything else? For me I feel it’s much more useful to learn HOW to approach these problems at a high level, rather than learning the specific mathematical or logical proof.

[@hellokitty7399]

SEND+MORE=MONEY =>SERD+MONE=MONEY =>SERD+MONE=10MONE+Y =>SERD=9MONE+Y

[@abhijeettalukdar5004]

MONE is not multiplied by 10 bhuh...... 😑

[@sultanelshirazy8229]

@@abhijeettalukdar5004 of course it is

[@sultanelshirazy8229]

@@abhijeettalukdar5004 just think of MONEY as a 5 digit number, like 12345. 12345 can be rewritten as 10 × 1234 + 5, am I right?

[@abhijeettalukdar5004]

@@sultanelshirazy8229 woah I got it , soory bro

[@sultanelshirazy8229]

@@abhijeettalukdar5004 yeah no problem, i do that too sometimes ☺️

[@sonairys929]

Finally A math problem on MYD channel that i can solve.

[@margaretcorfield9891]

Took me 5 mins. I was given one like this when I was about 8, to keep me busy at school. It took me less than half an hour, and my teacher was not happy that I'd solved it. The one I was given was CROSS+ROADS= DANGER It taught me a lot about numbers though, and encouraged a lifelong love of puzzles of all kinds.

[@Double-Negative]

OK. we all know the max you can carry in an addition problem like this is 1, so M=1. In order to carry the last digit, S=8 or 9. By this reasoning, O is either 0 or 1. M is already 1, so O=0. E+O=N implies something was carried into that, so N+R>10. E cannot be 9 because then E+O=N would mean N=0 or 1(a contradiction). This also proves nothing was carried into the first decimal place, so S=9. N+R=E+10 and N=E+1 implies R=9. This is a contradiction (S is already 9) so there is a carry there too, meaning N+R+1=E+10 and N=E+1 so R+1=9 and R=8. Also, D+E>11. D can't be 9 or 8, so E has to be at least 5. if we assume E=5, D can be 7, causing Y=2 and N=6. if we assume E=6, D cannot also be 6, meaning D=7, causing Y=3 and N=7=D (contradiction) E=7+ is ruled out because N=E+1 and the digits 8 and 9 are taken. E=5. the solution is 9567 +1085 =10652

[@WarmWeatherGuy]

If you let m be zero there are 24 more solutions.

[@simplebutpowerful]

I'm starting to look forward to Presh's classic tagline, "Did you figure it out?" now that I'm 'figuring it out' more frequently. Lol

[@fmaylinch]

What helped me was replacing N with E+1 in the equation. That way I saw: column 3: E + 0 = E+1 (This implies there's carry from column 2) Actually, it's also evident that column 2 has a carry when we see this: column 4: E+1 + R = E (The result is one less so R is big and causes overflow; actually R is 9 or it's 8 plus carry, which is the case here)

[@unformedvoid2223]

Figured it out! Spent at least 30 minutes for the solution, but solved it. It was fun, thanks :)

[@mithical1938]

I solved it! Took me 15-20 min in my head and I just wrote down the solved digits. Solved it exactly the same way too.

[@dw2843]

k

[@saurabhshukla2423]

This is not only one problem...you can choose any random word and then solve it...This problem is belongs to my course Artificial Intelligence... it's called cryptarithmatic problem...

[@priyeshajha9820]

I was so happy when I got this!

[@sciencetube4574]

Since M != 0, M = 1 (other carries are impossible to achieve with two numbers). So S + 1 (+ carry) = 10 (can't be 11 because of unique digits), meaning O = 0. This means either S = 8 and E = 9 and a carry from the previous addition or S = 9 and no carry from the third column. But if E = 9, then N + R (+ carry) = 19, which is impossible unless they're both 9. So S = 9. Furthermore, N = E + 1, since the numbers in the third column are distinct and N = E + 0 (+ carry), meaning there must be a carry and hence N = E + 1. Plugging this all into an equation gives 100E + 10N + 10R + D + E = 100N + 10E + Y 91E + 10R + D = 90N + Y, which, given that N = E + 1, resolves to E + D + 10R - Y = 90. The only possible solution for this is if R = 8 (since 9 is already taken) and E + D = 10 + Y. Now, the last question is, what is E + D. Y cannot be 0 or 1, and it also cannot be 3, since then E and D would have to be 6 and 7, respectively, leaving no room for N = E + 1, because 6, 7, 8, 9 are all taken. So Y = 2, E + D = 12. There is only one way to make 12 by summing two distinct digits without using 8 or 9, which is 5 + 7. Since N = E + 1, this means E = 5, N = 6 and D = 7. So the solution to SEND + MORE = MONEY is 9567 + 1085 = 10652, which is indeed a correct equation. I'll see if the video has a more elegant solution, but at least this solves it.

[@camerongray7767]

I tried, I got the same first 4 steps as you but couldn’t figure out the rest

[@Misteribel]

By en large I used the same approach, except that I used N=E+1 to find R=8, then deduced that D=E+2 which only allows E=5, D=7. Different routes, same method. Though I’m a little surprised it took you half an hour, I’m sure you just said that make us all feel good with our 10 min solutions 😄.

[@AnishWagle]

This is problem which I learned in computer engineering in 6th Shem in AI

[@whydoyouneedmyname6007]

I am in junior highschool and i solved it

[@chinareds54]

9567+1085=10652. If M0, then it must be 1 because there's no way to get a higher value from adding two four-digit numbers. This means S must be 8 or 9 in order to add to 1 and achieve a carry. The only possibilities for O are 0 and 1, but 1 is already taken by M, so O=0. Now you can either achieve that with 8+1+carry, or 9+1. But if E+0+carry > 10, then N would have to be 0 or 1 and those are taken, so we conclude that S=9 and E+0+carry= N, or N=E+1. But the fourth column also tells us that N+R+potential carry=1E=1[N-1]. The only thing we can add to fit this pattern is N+9=1[N-1]. 9 is taken, so R must be 8 and D+E must > 10 (so that N+R+carry=1E). Using the remaining numbers, we can brute-force the pairs for E-N, 6-7, 5-6, 4-5, 3-4, 2-3. 6-7 works for columns 3 and 4, but we're left with D+6=1Y with the highest remaining unused digit as 5. But then Y=1 which is taken by M. So we try 5-6. This works for columns 3 and 4, and if we make D=7, then D+E=7+5=12. Y=2. This is a solution. If we try 4-5, we end up with the maximum value we can make D+E to be 7+4=11. Y=1, which is already taken. 3-4 would yield Y=0 which is also taken. 2-3 doesn't allow us to make D+E>10. So our final answer is M=1, O=0, S=9, R=8, E=5, N=6, D=7, Y=2. 9567+1085=10652.

[@chinareds54]

Note: I numbered the columns from left to right for my explanation, which is reverse of Presh's numbering.

[@briansun6336]

Not very hard but a lot of logic is going on here. took me around 15 min. I find the geometry questions are hard for me.

[@billy.7113]

You are not alone. For hard geometry problems, more often than not, extra time won't help at all. You will be stuck for days unless you get a hint.

[@monnito3200]

I got another solution. What's wrong: Start by "send" + "more" = "money" From "more" we know m = 1000*(some digit) And in "money" it's 10000*(same digit) Doing the same thing on o,n and e And also y = 1*(digit) Then start simplifying : (Here's the process) "Send" + "more" = 10m + 10o + 10n + 10e + y Divide by 10: "Send"/10 + "more"/10 = m + o + n + e + y/10 Multiply by 10/"more" to get: "Send"/"more" = (10(m+o+n+e)+y)/"more So finally. "Send" = 10(m+o+n+e)+y Which is the value of "money"

[@rubixman7x7]

I figured out R by creating a table of possible E's and N's (2

[@rickyranggasaputra7926]

Next question S E N D M O R E ------------- + N U D E S

[@zdrastvutye]

it is valid because "sendmorude"has 10 different letters

[@mehdimabed4125]

Hi, I managed to solve the puzzle with an other method which much less elegant but quite funny : since abcd + defg = abfd + decg, the problem can be written as SERD + MONE = MONEY and if we put A = SERD and B = MONE, we have A + B = 10*B + Y => 9*B + Y = A with the last digit of B equal to the second one of A (this is the same letter E). Moreover, D + E

[@isaacchua0405]

All I got was that M=1

[@drjoephd]

This originally appeared as a puzzler on the ‘Car Talk’ radio show on public radio 13 or 14 years ago.

[@ssdfgsdfgsdfg1795]

not its original format, but yes, it was on Car Talk. Man, I miss that show :)

[@GyrlBlaque]

You lost me at saying hey. 😁

[@derpymule7977]

I was fairly certain I could solve this fairly easily, so I challenged myself to try solving it as quickly as possible without pen and paper, and I managed to figure it out all in my head in about 5 minutes (though I got lucky since I used trial and error for E,N,D,Y and E=5 was the first one I tried)

[@Officer-kd6]

isn't trail and error considered hack xD?

[@urvishbhavsar8910]

Hey.. I don't believe I solved it..I was going to see the video.. But when you said that we should really try it then u changed my mind and started solving it... And in 5 minutes I solved it without any paper or pen... Really I'm not telling lie...I was so happy when I solved the puzzle... I want to give thanks to Mr presh talwalker to encourage me and I solved it in the same way as you did...I still not believe that I have solved... Once again thanks to encourage me to solve this problem... And sorry for my bad English...

[@sonpham3438]

urvish bhavsar, would you mind sharing us ur thougths

[@Tyrian3k]

This is how I approached it: M can only be 1 since you can only carry over at most 1 from a sum of two numbers. => *M=1* O could only be 0 or 1 at most, but since 1 is already taken, *O=0.* E+O must be less than 10 since N would otherwise be 0, which is already taken, so there can't be a carry over. Therefore *S=9.* For E to not be equal to N, there must be a carry over from N+R=E and therefore N=E+1. Since N=E+1, N+R=E can only be true if *R=8* and there is a carry over from D+E=Y. Since Y can't be 0 or 1, D+E must be 12+ and since N=E+1, the only way to achieve that is if D is the highest untaken number, so *D=7* E must also be the highest number it can be, With 7,8 and 9 taken and N=E+1, *E=5.* Therefore D+E=12 which means *Y=2.* N=E+1, so *N=6* So it is 9567 +1085 = 10652

[@Armando-zl5eq]

Subtitulos al español por favor

[@N8570E]

Did I figure it out? No way! But I enjoyed following your solution, very slowly and with many rewinds. Thank you for your efforts. May you and yours stay well and prosper.

[@aditiagrawal8750]

Only a genius could solve it? I solved it. Does it mean I am a genius??

[@nagindersood9695]

No, you are just an exception 😆😆

[@Clarissa_M]

Okay, but if I was forced against my will to recommend a MindYourDecisions video for someone to try and figure out by themselves, this would absolutely be the one! I wasn't planning to try and solve this; I just thought the ultimatum on the thumbnail was funny. But the comments were so positive and I gave it a try. It was so fun and quite satisfying!!

[@thesentientneuron6550]

This problem was in my 7th grade math textbook

[@Nihcasak]

Are you from India (Maharashtra)? I also had the same problem in seventh grade maths textbook. 12 year old myself figured it out using trial and error method. And when I learned programming, I actually wrote c++ program to check out all possible solutions. The logical method shown in this video is very intresting!. Please continue making videos like this!. Thank you!.

[@Everything_in_everyone_was]

This is a classic cryptarithm puzzle where each letter represents a distinct digit. To solve the puzzle, we need to use some logic and trial and error to assign a unique digit to each letter in a way that satisfies the equation. One approach is to start with the leftmost letter and work our way towards the right. Since the sum of the two numbers has to be a four-digit number, we know that M must be either 1 or 2 (because if M=3 or higher, the sum would be at least 3000+3000=6000, which is a five-digit number). Let's assume that M=1. Then, the equation becomes: 1ORE +1ORE ------ MONEY Since E+ E = Y, we know that E cannot be 0, 5, or higher. Let's assume that E=6 (we can check later whether this assumption is correct or not). Then, the equation becomes: 1OR6 +1OR6 ------ MO6Y Since 1+1=2, we know that O+O+1=Y, which means that Y must be an even number. The only even digit that hasn't been used yet is 8, so we can assume that Y=8. Then, the equation becomes: 1OR6 +1OR6 ------ MO68 Since R+R+1=8, we know that R=3. Then, the equation becomes: 1O36 +1O36 ------ MO68 Since O+O=6, we know that O=3. But this violates the rule that each letter must represent a distinct digit. Therefore, our assumption that M=1 must be wrong. Let's try the other possibility, which is M=2. Then, the equation becomes: 2ORE +2ORE ------ MONEY Since E+E= Y, we know that E cannot be 0, 5, or higher. Let's assume that E=4. Then, the equation becomes: 2OR4 +2OR4 ------ MO4Y Since 2+2=4, we know that O+O+1=Y, which means that Y must be an odd number. The only odd digit that hasn't been used yet is 5, so we can assume that Y=5. Then, the equation becomes: 2OR4 +2OR4 ------ MO45 Since R+R+1=9, we know that R=4. Then, the equation becomes: 2O44 +2O44 ------ MO45 Since O+O=8, we know that O=4. But this violates the rule that each letter must represent a distinct digit. Therefore, there is no solution with M=2 either. Therefore, there is no solution to this cryptarithm puzzle with each letter representing a distinct digit.

[@boerbieh]

We can... MATH... the world a better place... :)

[@tannhauser7584]

Our algebra teacher gave us a similar problem. HOCUS + POCUS = PRESTO. Nobody solved it that day. Years later, I remembered it and started working on it again. It took several hours, but I found a valid solution. Sadly, it has been lost to history and I don't feel like doing it again.

[@KommentarSpaltenKrieger]

Lol, took me about 2 hours. I came up with the first few digits logically and then it just boiled down to assuming carryovers and trial and error until it all fit together.

[@salmanel-farsi3744]

To persevere for 2 hours and not give up. That is great to hear. A lot of people abandon after such a short period of time. Even if you applied a combination of logical deduction and brute force (trial and error), the main point is you stuck with it and I am sure you gained a lot of insights about carrying over along the way.

[@ironcladranchandforge7292]

Yes, I did figure it out. First thing I did was grab some paper and wrote down some logical conclusions, such as: N has to be larger than 1. E has to be less than N. M has to be 1. O has to be 0. N + R has to be larger than 9. S +M has to be larger than 9. N + R, one or both have to be larger than 5..... Once I came to these logical conclusions it was easy to figure out. Took me about 30 minutes as well.

[@abhijeettalukdar5004]

I wonder why boys don't get our signs Their signs:

[@MrFrmartin]

Got most of it, then forgot to get back to finish and lost my work, but I really like working through these. Helps I went through some similar ones

[@notsean9862]

Put the speed at 1.25x and he sounds normal

[@arthurgraton7165]

You. Are a genius.

[@bartmeijer1954]

There is another puzzle much like this one: THIS + ISA + GREAT + TIME = WASTER. Each letter stands for a unique digit, and in each word the last letter corresponds to the digit in the 1's position. I solved it as a teenager, but I don't remember the answer.

[@kristiankember8973]

damm it again I made a false assumption. I thought no digit could be 0 :(

[@OneWeirdDude]

If there are multiple digits in a number, the first cannot be 0; others can be zero.

[@kristiankember8973]

I knew that it was the little message at the bottom that my random assumption was that he ment all the letters couldn't be 0

[@robinlindgren6429]

some notation: r(1) means remainder from 1s column, r(10) means remainder from 10s column, etc, r(1)? means we have not confirmed or refuted the existence of this remainder, r(1) is TRUE or r(1) is FALSE means the 1s digit did or did not produce remainder. SEND+MORE=MONEY since M is not 0, M must be 1, since no 2 numbers on the 1000s digit, even with a r(100), can produce a remainder of 2, M=1 SEND+1ORE=1ONEY we therefore have that S+1+r(100)?=1O, which means that S must be either 8 or 9, and O must be either 0 or 1, since it cannot be 1 (it must be different from M), it is 0. SEND+10RE=10NEY in the 100s digit, we have E+0+r(10)?=N, if r(10) is FALSE, then this would be E+0=E, therefore r(10) must be TRUE, giving us E+1=N+r(100)?. if E+1 produces a remainder, then E must be 9, which it cannot be because S is 9, therefore r(100) is FALSE. therefore, we have in the 1000s digit: S+1=10 : S=9 9END+10RE=10NEY in the 100s digit we have E+1=N, and in the 10s digit we have N+R+r(1)?=E+r(10)?, insert N=E+1 and we get E+1+R+r(1)?=E+r(10)?, simplify to 1+R+r(1)?=r(10)? since adding something non-negative to 1 cannot produce 0, the right side of this equation must be positive, therefore, r(10) is TRUE, after simplification we get: R+r(1)?=9 since R cannot be 9 (S is 9) that means R=8 and r(1) is TRUE 9END+108E=10NEY lets bring our attention to what Y is, we know that r(1) is TRUE, therefore D+E=10+Y, Y cannot be 0 or 1, if Y is greater than 2 then D+E must be at least 13. since 9 and 8 are already taken, we would need to use 7+6, this is however impossible since that would leave N=E+1 as either 8 or 7, both of which would already be taken, therefore Y must be 2. 9END+108E=10NE2 therefore, D+E=12, this cannot be 9+3 (9 is taken), it cannot be 8+4 (8 is taken), it cannot be 6+6 (both must be different), therefore it is 7+5, E cannot be 7 since that would leave N=E+1 as 8, which is taken, therefore E=5, N=6, D=7 9567+1085=10652 S=9 E=5 N=6 D=7 M=1 O=0 R=8 Y=2

[@jbtechcon7434]

I did this whole thing in my head in under a minute on the toilet. How is it even possible that I'm single?

[@urvishbhavsar8910]

JBTechCon no you're not alone

[@samarthmehta8732]

Maybe, because you're a liar.

[@safiyahmohammed3848]

I wanna believe

[@sonpham3438]

JBTechCon, you probably shat all the humanity out of your body

[@shrihanshetty8859]

Hey I got another solution don't know whether it's legible but here goes- D=0 E=6 N=9 R=7 O=2 M=1 so you get 12966

[@distmuchin4564]

Wouldn't E and Y be the same digit then? Each letter must correspond to a distinct digit.

[@Kali-bs7oj]

Idk if there’s any merit to this strategy but I changed it to be written S E R D + M O N E M O N E Y Because the MONE is repeated and kind of has a pattern

[@clarawoodman9331]

Those kind of puzzles are my favourite, but the best one I've done was in Italy it went like this: • • A × B B C = C A D A + : - A E F + C F = A D D = = = A D G + C E = C H H C You have to solve it vertically and horizontally! (I swear it's fun to solve it)

[@ValehFarzaliyev]

the problem doesn't specify which number system to use. There are seven distinct letters in the puzzle, so we can use any number system with at least 7 digits.For example, base-8 number system. I found that 6342 + 1073 = 10435 is also solution

[@hans-rudigerdr.zimmermann5672]

Hi Presh, yes this is a good problem. I solved it in 25 minutes. It is 9567 + 1085 = 10653. It is a little bit tricky. . I explain: first observe the M must be 1, since 0 is not allowed. Then you see that O must be 0 since S has only 1 digit. S can be 9 or 8, but 8 is not possible since E plus 0 can not sum up to a dobble digit with the first digit 1. So S = 9 . Now E + 0 + 1 = N. We need the 1 if not we would get E= N. So E+1= N. It follows that R = 8 since N +1 + R must have as the second digit E. Obviously R can not be 9 because S= 9. Now N+ 9 has as second digit E and D + E has as second digit Y. But this can only be resolved if N = 6 , E = 5. It follows that D = 7 and Y = 2. ARRIVING to the correct equation 9567 + 1085 = 10653

[@ve2zzz]

Woow Superb !!! I resolved it the same way Presh did it !!! This puzzle REALLY deserves you to pause the video, get paper and pencil and spend 30 minutes to resolve it !! Thank you Presh ! Hint: Pay attention to carries.

[@keithbar19741974]

I am sensing a shortcut here that I have not fully explored. If you swap the "N" & "R" in the above which you can do as it is addition you get "SERD" + "MONE" = "MONEY". In this case you can see that "MONE" is just the first 4 letters on "MONEY" and in fact the answer shows that. Is there a shortcut way to change this into subtraction of shifted money?

[@marti85cr]

This is very easy M = 1, since 9+8+1(possible carry) = 18, the max combination one can get in a column of this exercise. since M = 1 S + 1 + [carry] = 1O, here there are two options out of three combinations a) S + 1 + 1 = 10 b) S + 1 + 1 = 11 (O can't be 1, because M is already 1) c) S + 1 + 0 = 10 so O = 0, and S is 8 or 9 E + 0 + [carry]= N, carry must be 1, since E + 0 = E, and this column doesn't produce a carry, since N cannot be equal to 0 nor 1, thus S=9 and N=E+1 N + R + [carry]= 1E a) N + R + 1 = 1E E +1 + R + 1 = 10+E R+2= 10, R=8 b) N + R + 0 = 1E E +1 + R + 0 = 10+E R+1= 10, R=9, but S is already 9, so R= 8, and there's carry from the previous column N+9 = 10 +E D + E = 10+Y, Y must be higher than 2, E can't be 7 (N should be 8 but R is already 8), so E should be 5 (can't be 4, because then D = 7 o 6, and Y would be 1 or 0, can't be neither 6, because D would be 5, and D+E = 11 (Y = 1 is not possible) So E = 5, N = 6, D = 7, D+E=12 and Y = 2 9567 +1085 ====== 10652

[@loganv0410]

Slightly different logical steps but same answer That's one of the things that drew me to math: in the midst of the social uncertainties of the 1960's in the South, plus the uncertainties of puberty, math was an internally consistent island of calm.

[@gnmurthy1954]

During college days, I came across a similar problem which is FORTY+TEN+TEN=SIXTY. All the digits are used and are non repetitive.

[@takdanaicheepsaritdipong9381]

I finally solved one of your puzzle. Since I have watched your channel for years, I solve at most 3 questions lol. Thank you so much Presh

[@aadarshsavaikar4337]

I have a similar question: ABC-CBA=CAB, and given A>B>C & {A,B,C are not 0.}

[@YOM2_UB]

O = 0, M = 1, Y = 2, E = 5, N = 6, D = 7, R = 8, S = 9. Following is my logic used to reach this conclusion: M ≠ 0, so the addition of two 4-digit numbers has a 5-digit sum, which is only possible if the leading digit of the sum is 1. Thus M = 1. S + M + c = 10 + O, where c is the carry digit (1 or 0 depending on the sum of digits one space right and their carry digit). Since M = 1: S + c = 9 + O. Since S is a single digit, this means S must be either 8 or 9. Let's look at the possibility of S = 8. The carry digit would then need to be 1 and O would need to be 0. Looking at the next column though, E + O + c = 10 + N, if O = 0, E would need to be the 9 with a carry digit of 1 to work, but it couldn't then have a sum of 10 + N, as N would need to be 0, which would be taken by O. Thus, S = 9. S + M + c = 10 + O. Plug in S = 9 and M = 1: 9 + 1 + c = 10 + O --> c = O. If c = 1, O = 1, but M = 1 and digits can't repeat, so O = 0, and E + O + c < 10. E + O + c = N. As O = 0, if c = 0 this would result in E = N, but E ≠ N so c = 1, and N = E + 1. N + R + c = 10 + E. Plug in N = E + 1: E + R + c = 9 + E. S = 9, so it must be that R = 8 and c = 1. D + E = 10 + Y (no c as it's the rightmost column). The remaining possible digits are 2, 3, 4, 5, 6, and 7. There are then two possible solutions for D + E = 10 + Y: [Y = 2, {D, E} = {5, 7}] and [Y = 3, (D, E} = {7, 6}]. Of these, we can rule out the solution where Y = 3 by taking into account that N = E+1. If the solution were true, N would need to be either 7 or 8, and both would have been taken. Thus Y = 2. {D, E} = {5, 7}, N = E + 1. If E = 7, then N = 8 but 8 is taken, so E = 5. Thus N = 6 and D = 7.

[@esotericVideos]

What helped me is that I made a grid with each row being a letter and each possible number being a column, then just put X's everywhere that was impossible.

[@aneeshgupta6941]

Awesome question.. btw.. I was able to solve this...😍😍 Really feeling proud after solving this problem...